Does the circuit configuration (to measure inductance using 'pulse method') you meant is the one shown in figure below?Mr.Cool said:by the way, the IGBT circuit i described should be a half bridge (two igbts) only because the upper igbt has a body diode that can use to disipate the stored energy in the inductor during pulse OFF period. and you should use a large cap across the half bridge because only caps can discharge that kind of energy in short period of time. a power supply can't do that..
Mr.Cool
I'm not sure about the way to get dI. Do you mean find the dI = Ipeak from the scope display, or need some integration to find dI from the signal captured by the scope?Mr.Cool said:now you can calculate your inductance via:
V = L * dI/dT
where V is the bus voltage and dT is your pulse width. dI is found from the current sensor output. solve for L.
The pulse width should be choosen in such a way that the di/dt is linear. Pls correct me if I'm wrong.Mr.Cool said:what is a good pulse width to use?? this is a good question. i'll let you think about it for a while..
picotube said:The pulse width is 50us.
Therefore,
L = V*dt/di = 4V * 50us / 19A = 10.5uH (close to 10uH)
The pulse width should be choosen in such a way that the di/dt is linear. Pls correct me if I'm wrong.Mr.Cool said:what is a good pulse width to use?? this is a good question. i'll let you think about it for a while..
afonso said:When doing this experiment in a real circuit, won't do it whit a duty cycle so high. Or the core wil saturate and burn the circuit!
As I don't really understand, sorry to ask the following questions:Mr.Cool said:if you run a continuous pulsing circuit (duty cycle) then you run the risk of over heating due to inductor/transformer saturation.
3) Why when the core saturates, the flux linkage = 0 ? If I'm not mistaken, I learned before that when the operation of an inductor or a transformer reaches the saturation region, the flux linkage is no more increasing linearly with current anymore. Or, in other words, the flux linkage does not increase as the same rate as the current increases. But the flux linkage is not zero, right? Probably the rate of change of flux linkage when the core/inductor/transformer is saturated is 'small', so that the induced voltage ( Lens Law, or V = d(flux-linkage)/dt ) is not large enough and therefore the current increases rapidly. I studied this few years ago so some of my memory cells might be damaged already and therefore remember wrongly. I'm very confusing now. Pls advise and pls correct me if I'm wrong.Mr.Cool said:when the core saturates the flux linkage = 0 and your current has rises very fast as it is limited only by the wire resistance.
Agree. When core is saturated, permeability of the core reduces (see attached figure), and hence the inductance of the inductor/transformer reduces as inductance is proportional to permeability.Mr.Cool said:when the core saturates the permiability is reduced to very small number. since inductance is proportional to relative permiability then the inductance reduces also, very close to zero. if you are using a core in the inductor/transformer design it is because you are after large energy storage, usually large inductance values too. thus, when i say inductance zero, i mean in comparison to the original value it is orders of magnitude smaller.
If I'm not mistaken, at any point on the curve, permeability is given by the ratio (B/H) and not by dB/dH (the slope of the curve), right? ( Pls refer http://www.ee.surrey.ac.uk/Workshop/advice/coils/mu/#mu )Mr.Cool said:think of the B-H curve. as you increase in magnetization strength the flux levels off and doesn't increase anymore. since the permiability is the slope of the B-H curve, and the slope is zero (or close to zero) at the saturation point then it follows that the permiability is zero. thus inductance falls away.
I see the picture now. I get your points now. Great !Mr.Cool said:V = L*di/Dt
here, V stays the same, dT stays the same.. but L is falling. to compensate for this then the only variable left is dI - which must rise.
this is the begining of the end because as current increases it forces the saturation even more until even the voltage begins to breakdown.
it all comes down to seeing how V= L*di/dT is formed
L = u * n^2*A / l_m
when flux density can no longer increase with increasing current then the rate of change of flux density falls away to zero.
this is Faraday's law you mention --> v = n*A*(dB_sat/dT) = 0
since v = 0 you have a short circuit... the current is now limited only by series resistance.
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